Year: 2023
Paper: 3
Question Number: 5
Course: UFM Additional Further Pure
Section: Number Theory
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Show that if
\[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,,\]
then $(2x - 7)(y - 7) = 49$.
By considering the factors of $49$, find all the pairs of positive integers $x$ and $y$ such that
\[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,.\]
\item Let $p$ and $q$ be prime numbers such that
\[p^2 + pq + q^2 = n^2\]
where $n$ is a positive integer. Show that
\[(p + q + n)(p + q - n) = pq\]
and hence explain why $p + q = n + 1$.
Hence find the possible values of $p$ and $q$.
\item Let $p$ and $q$ be positive and
\[p^3 + q^3 + 3pq^2 = n^3\,.\]
Show that $p + q - n < p$ and $p + q - n < q$.
Show that there are no prime numbers $p$ and $q$ such that $p^3 + q^3 + 3pq^2$ is the cube of an integer.
\end{questionparts}
Whilst this was the most popular question, it was only the sixth most successful with a mean mark of a little under 9/20. Many of the candidates made substantial attempts at parts (i) and (ii) but found it more challenging to make progress with part (iii). Two common general errors were lack of precision when handling inequalities, and working backwards from a required result without demonstrating that the logic was reversible. In part (i), a small number of candidates rearranged the first equation to remove denominators then wrote the required result without adequate intermediate steps of working. There were a very small number of arithmetic errors when finding the pairs of x and y. In (ii) many candidates commented that as p and q were prime then the only possible factors of pq were 1, p, q and pq and went on to test each of these as possible values for p + q + n. Many candidates were able to form a relevant equation involving p and q and whilst most factorised it, similarly to part (i), a small number attempted alternative approaches. The most successful of these was to write p as a function of q and rewrite the improper fraction to see that q − 2 must divide 3. A small number of candidates spotted that p and q were the solutions to the quadratic equation t² − (n + 1)t + (2n + 1) = 0 but from here few were able to fully justify that the only solutions for (p, q) came from n = 7. The first two results of part (iii) caused much confusion. Relatively few candidates realised at the start of their attempts that these were equivalent to q < n and p < n. Those who did recognise this completed part (iii) with relative ease. For the second part, a pleasing number of candidates realised that (p + q)³ would be a useful expression to consider and those who did usually managed to get to the difference of two cubes expression necessary to make progress. Some candidates were unsure where to go next but a good number realised the importance of the printed inequalities and correctly deduced that p + q − n must be 1 or 3. From here candidates often managed to rule out one case but ruling out both successfully was relatively rare.